- A smooth horizontal table has a vertical post fixed to it which has the form of a circular cylinder of radius $\displaystyle a$.
- A light inextensible string is wound around the base of the post ( so that it does not slip ) and its free end of the string is attached to a particle that can slide on the table.
- Initially the unwound part of the string is taut and of length $\displaystyle 4a/3$.
- The particle is then projected horizontally at right angles to the string so that the string winds itself on to the post.
- How long does it take for the particle to hit the post ?.
You may make use of the formula
$$ \int\sqrt{\, 1 + \phi^{2}\,}\,\mathrm{d}\phi = \frac{1}{2}\,\phi\,\sqrt{\, 1 + \phi^{2}} + \frac{1}{2}\sinh^{-1}\left(\phi\right)\tag A$$
Below is solution recommended by professor, however I would like to make use of the formula $(A)$ suggested above?.
$\dfrac{mv^2}{2} = E$
Let $\theta$ be the angle between the direction of the free string at time $t$ and its initial direction, as shown in Figure. Since the length of the free string at time $t$ is $b-a\theta$, it follows that $v=(b-a\theta)\dot\theta$. On using the initial condition $v=u$ when $t=0$, the energy conservation equation becomes:
$(b-a\theta)\dot\theta=u$
Solving differential equation
$$\int_0^{\frac{b}{a}}(b-a\theta)d\theta = u \int_0^{\tau}dt$$
\begin{align}\tau &= \frac{1}{u} \int_0^{\frac{b}{a}}(b-a\theta)d\theta\\&= \frac{1}{u}[b\theta - \frac{1}{2} a \theta^2]_0^{b/a} \\&= \frac{b^2}{2au}\end{align}
What sort of solution to be done to make use of the formula (A)?